19 23 29 triangle

Acute scalene triangle.

Sides: a = 19   b = 23   c = 29

Area: T = 218.1566337291
Perimeter: p = 71
Semiperimeter: s = 35.5

Angle ∠ A = α = 40.85546193501° = 40°51'17″ = 0.71330476223 rad
Angle ∠ B = β = 52.35993005849° = 52°21'33″ = 0.91438421892 rad
Angle ∠ C = γ = 86.7866080065° = 86°47'10″ = 1.5154702842 rad

Height: ha = 22.96438249779
Height: hb = 18.97701162861
Height: hc = 15.04552646407

Median: ma = 24.38774967965
Median: mb = 21.65106350946
Median: mc = 15.3221553446

Inradius: r = 6.14552489378
Circumradius: R = 14.52328419186

Vertex coordinates: A[29; 0] B[0; 0] C[11.60334482759; 15.04552646407]
Centroid: CG[13.53444827586; 5.01550882136]
Coordinates of the circumscribed circle: U[14.5; 0.81442096728]
Coordinates of the inscribed circle: I[12.5; 6.14552489378]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 139.145538065° = 139°8'43″ = 0.71330476223 rad
∠ B' = β' = 127.6410699415° = 127°38'27″ = 0.91438421892 rad
∠ C' = γ' = 93.2143919935° = 93°12'50″ = 1.5154702842 rad

Calculate another triangle




How did we calculate this triangle?

Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.

a = 19 ; ; b = 23 ; ; c = 29 ; ;

1. The triangle circumference is the sum of the lengths of its three sides

p = a+b+c = 19+23+29 = 71 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 71 }{ 2 } = 35.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 35.5 * (35.5-19)(35.5-23)(35.5-29) } ; ; T = sqrt{ 47592.19 } = 218.16 ; ;

4. Calculate the heights of the triangle from its area.

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 218.16 }{ 19 } = 22.96 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 218.16 }{ 23 } = 18.97 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 218.16 }{ 29 } = 15.05 ; ;

5. Calculation of the inner angles of the triangle using a Law of Cosines

a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 19**2-23**2-29**2 }{ 2 * 23 * 29 } ) = 40° 51'17" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-19**2-29**2 }{ 2 * 19 * 29 } ) = 52° 21'33" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-19**2-23**2 }{ 2 * 23 * 19 } ) = 86° 47'10" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 218.16 }{ 35.5 } = 6.15 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 19 }{ 2 * sin 40° 51'17" } = 14.52 ; ;




Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.